Can you simplify cos(x)cos(2x)cos(4x)cos(8x)cos(16x) ... cos(2^n x) ?,

1 Answer
Feb 8, 2018

sin(2^(n+1x))/(2^(n+1)sin(x))

Explanation:

We want to simplify

cos(x)cos(2x)cos(4x)cos(8x)...cos(2^nx)

The trick is to use the double-angle identity repeatedly

  • sin(2x)=2cos(x)sin(x)

Consider the following example

sin(16x)=2cos(8x)sin(8x)

=2^2cos(8x)cos(4x)sin(4x)

=2^3cos(8x)cos(4x)cos(2x)sin(2x)

=2^4cos(8x)cos(4x)cos(2x)cos(x)sin(x)

=>cos(x)cos(2x)cos(4x)cos(8x)=sin(16x)/(2^4sin(x))

If we generalize it

sin(2^(n+1)x)=2cos(2^nx)sin(2^nx)

=2^2cos(2^(n)x)cos(2^(n-1)x)sin(2^(n-1)x)

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=2^(n)cos(2^(n)x)cos(2^(n-1)x)...cos(4x)cos(2x)sin(2x)

=2^(n+1)cos(2^(n)x)cos(2^(n-1)x)...cos(2x)cos(x)sin(x)

=>cos(x)cos(2x)cos(4x)cos(8x)...cos(2^nx)=sin(2^(n+1)x)/(2^(n+1)sin(x))